## Order with Respect to Each Reactant

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

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Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Order with Respect to Each Reactant

I have a general question about the order with respect to each reactant. I was wondering, if you solve for the order with respect to a reactant and get a number like 1.89 or 2.11, can you round up or round down, respectively, to 2? For example, in problem 15.19 (a), when solving for the order with respect to B, I set the ratios (3.02^b / 1.25) and (457 / 50.8) equal to each other and got 3.02^b = 7.2989. I solved for b and got 1.80. Can you round this number up to 2 and say that B is second order?

Isabella Sanzi 2E
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

### Re: Order with Respect to Each Reactant

In the example you used, I just wanted to point out that for this problem that you may have made a small error because I got an integer for this one. I used reactions 1 and 3 because [B] changes while [A] and [C] remain the same. Since 3.02/1.25 = 2.416 and 50.8/8.7 = 5.84 (not 457, that would be from reaction 4, which changes the concentration of both [B] and [C]), this works out conveniently because 2.416^2 is extremely close to 5.83, which would indicate that this is a second order reaction. (This is the same principle as a reactant concentration doubling, and and the rate being multiplied by four so you have 2^2 = 4, indicating a second order reaction).
I do not know the exact answer to your general question, but I think that these problems usually will work out in ways that it is obvious what the order is (confirmation on this would be helpful though).

Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Re: Order with Respect to Each Reactant

Yeah, sorry. I meant the ratios (3.02^b / 1.25) and (50.8 / 8.7). I mixed up part (b) and part (c). Thank you!

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